### Subject: Pendulum: What is relationship of the T^2 vs. length?

Date: **Tue Dec 24 10:02:04 2002**

Posted by **Michael**

Grade level: **10-12**
School: **Life Sciences Secondary**

City: **New York** State/Province: **NY**
Country: **US**

Area of science: **Physics**

ID: **1040742124.Ph**

**Message:**

In a pendulum lab, one gets a straight line if he/she graphs the period-squared (T^2)
vs. the length of the string.

The relationship equation is T = 2*pi*SQRT(length/g).

If you graph the string length on the y-axis and the T^2 on the x-axis (which would
would technically be incorrect because the string length is the independant variable), the
slope's units are meters/sec^2.

Does this slope represent the value of g (acceleration due to gravity)? If so, then how
can it? I solved the equation T = 2*pi*SQRT(length/g) for "g."

g = length/(T/2*pi)^2

So how can one simply "exclude" the 2*pi factor and assume that the above slope
represents "g." In a lab book, I see a question: "What does the slope represent? [Hint:
solve your equation for length/T2.]" My answer would be that it represents nothing of
importance. Only graphing "length" vs. "(T/2*pi)^2" would represent "g," which would be
important.

Is my thinking off?

MRP

Re: Pendulum: What is relationship of the T^2 vs. length?

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